What Is the Correct Substitution for This Integral?

[ascky]

I've been trying to figure this out for a while, but I can't find the right substitution:

\int \sqrt{\frac{x-1}{x(x+1)}} dx

Anyone able to help me out? It's driving me crazy.

 

[Pseudo Statistic]

Well, Wolfram seems to think that it reduces to the sum of an elliptic integral of the first kind and second kind, furthermore involving complex numbers.

 

[ascky]

Oh, gee, I copied it wrong. How embarrassing. My apologies. How about

\int \frac{\sqrt{\frac{x-1}{x+1}}}{x^2} dx

Wolfram thinks it's arctan-ny.

 

[morson]

Make the substitution x = sin^2h(t), and the integral reduces to:

2\int \sqrt{sin^2h(t) - 1}\ dt

Edit: just saw your post. Oops.

 

[D H]

You get an ugly mess if you type that into Wolfram's Integrator directly. On the other hand, simplifying the radical leads to the much simpler form
\int\frac{\sqrt{x^2-1}}{x^2(x+1)}dx =<br /> -\left(\tan^{-1}\left(\frac1{\sqrt{x^2-1}}\right)+\frac{\sqrt{x^2-1}}x\right)

 

[morson]

In that case, make the substitution u^2 =\frac{x-1}{x+1}\.

So the integral becomes:

\int \frac{4u^2 du}{(1+u^2)^2}\

Looks like a further substitution of u = tan(t) or u = sinh(a) will do the trick. Or partial fractions.

By partial fractions, the integrand is: \frac{4}{1+u^2}\ - \frac{4}{(1+u^2)^2}\

 

[ascky]

I'm having trouble with the algebra manipulating that substitution to get it nicely as

\int \frac{4u^2}{(1+u^2)^2} du

Would it be too much trouble to explain how you did that? Of course from there, it's nice.

 

[morson]

u^2 = \frac{x-1}{x+1}\
= 1 - \frac{2}{x+1}\

Solving for (x + 1) gives:

x + 1 = \frac{2}{1 - u^2}\ --- (1)[/color]

So, (x + 1)^2 = \frac{4}{(1 - u^2)^2}\ --- (2)[/color]

Also, 2u du = \frac{2 dx}{(x+1)^2}\

^ You got that part right? So:

dx = u(x + 1)^2 du

From (2)[/color],

dx = \frac{4u du}{(1 - u^2)^2}\

So, the expression inside the square root of your original integral becomes u^2, so the whole square root becomes u. From (1)[/color]:

x = \frac{2 - (1 - u^2)}{1 - u^2}\

= \frac{1 + u^2}{1 - u^2}\

Therefore, x^2 = \frac{(1 + u^2)^2}{(1 - u^2)^2}\

So, the whole integral becomes:

\int \frac{ \frac{4u^2 du}{(1 - u^2)^2} }{ \frac{(1 + u^2)^2}{(1 - u^2)^2}\ } \

= \int \frac{4u^2 du}{(1 + u^2)^2}\

 

[ascky]

Ahhh I get it now. That was very clear.

Thanks very much for your help. Now I can sleep contentedly!

 

[i.mehrzad]

well i tried it by substituting the following:

x=\tan^2\t and then continued. It simplifies very well.

 

[i.mehrzad]

I never get my latex coding right.
x/=/tan^2t

 

[i.mehrzad]

well isn't the backslash to be for space,

x\=\tan^2\t

 

[i.mehrzad]

now i got it all wrong
x=\tan^2t

 

[i.mehrzad]

Now it is fine, sorry for my disobedience with so many posts.